Simplify the following expression and state the condition under which the simplification is valid. You can assume that $x \neq 0$. $k = \dfrac{8x^2 + 12x}{-3} \times \dfrac{2}{4(2x + 3)} $
Solution: When multiplying fractions, we multiply the numerators and the denominators. $k = \dfrac{ (8x^2 + 12x) \times 2 } { -3 \times 4(2x + 3) } $ $ k = \dfrac {2 \times 4x(2x + 3)} {-3 \times 4(2x + 3)} $ $ k = \dfrac{8x(2x + 3)}{-12(2x + 3)} $ We can cancel the $2x + 3$ so long as $2x + 3 \neq 0$ Therefore $x \neq -\dfrac{3}{2}$ $k = \dfrac{8x \cancel{(2x + 3})}{-12 \cancel{(2x + 3)}} = -\dfrac{8x}{12} = -\dfrac{2x}{3} $